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A261497 Number of necklaces with n white beads and 3*n black beads.

%I A261497 #17 Feb 16 2025 08:33:26
%S A261497 1,1,4,19,116,776,5620,42288,328756,2615104,21191904,174303163,
%T A261497 1451430692,12211799224,103655949072,886568153744,7633233556276,
%U A261497 66105170315084,575445692499952,5032380942945322,44191451788248416,389514699013012242,3444925385336301684
%N A261497 Number of necklaces with n white beads and 3*n black beads.
%H A261497 Alois P. Heinz, <a href="/A261497/b261497.txt">Table of n, a(n) for n = 0..1000</a>
%H A261497 F. Ruskey, <a href="http://combos.org/necklace">Necklaces, Lyndon words, De Bruijn sequences, etc.</a>
%H A261497 F. Ruskey, <a href="/A000011/a000011.pdf">Necklaces, Lyndon words, De Bruijn sequences, etc.</a> [Cached copy, with permission, pdf format only]
%H A261497 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Necklace.html">Necklace</a>
%H A261497 Wikipedia, <a href="https://en.wikipedia.org/wiki/Necklace_(combinatorics)">Necklace (combinatorics)</a>
%H A261497 <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%F A261497 a(n) = 1/(4*n) * Sum_{d|n} C(4*n/d,n/d) * A000010(d) for n>0, a(0) = 1.
%F A261497 a(n) ~ 2^(8*n-3/2) / (3^(3*n) * sqrt(3*Pi) * n^(3/2)). - _Vaclav Kotesovec_, Aug 22 2015
%p A261497 with(numtheory):
%p A261497 a:= n-> `if`(n=0, 1, add(binomial(4*n/d, n/d)
%p A261497              *phi(d), d=divisors(n))/(4*n)):
%p A261497 seq(a(n), n=0..25);
%Y A261497 Column k=3 of A261494.
%K A261497 nonn
%O A261497 0,3
%A A261497 _Alois P. Heinz_, Aug 21 2015